To find the future value of the investment when interest is compounded monthly, we use the formula:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Here,

• $P = \$4500$
• $r = 0.0415$ (annual interest rate)
• $n = 12$ (monthly compounding, so 12 times per year)
• $t = 3$ years

Substitute the values into the formula:

$A = 4500 \left(1 + \frac{0.0415}{12}\right)^{12 \cdot 3}$ $A = 4500 \left(1 + \frac{0.0415}{12}\right)^{36}$ $A = 4500 \left(1 + 0.00345833\right)^{36}$ $A = 4500 (1.00345833)^{36}$

Now, we calculate $(1.00345833)^{36}$:

$(1.00345833)^{36} \approx 1.129620$

So,

$A = 4500 \times 1.129620$ $A \approx 5083.29$

Therefore, the future value of the investment when interest is compounded monthly is $5083.29.

Would you like details or have any questions? Here are 5 questions for you to ask:

1. How do you calculate the future value with daily compounding?
2. What is the formula for future value with continuous compounding?
3. Can you explain the significance of different compounding intervals?
4. How does the compounding frequency affect the future value?
5. What are some practical applications of these compounding formulas?

Tip: When calculating compound interest, ensure you correctly adjust the interest rate and the number of compounding periods to match the compounding frequency.